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The remaining procedure is the same as before:
det A T A
2 n
det ( K )
α n + 1
(6.35)
n
ζ
det A T A
2 n
2
2
det A T A
b
α n + 1
(6.36)
2 n + 1
ζ
n
(
1
ζ )
ζ
n
2
2
2 ( 1 ζ ) =
b
γ OLS
1 ζ
α n + 1
(6.37)
Considering also that
σ 2
σ 2
i
2
i 1
2
α i
i = 2, ... , n
(6.38)
ζ
ζ
it can be deduced that for
tending to zero (OLS), all the eigenvalues of matrix
K but the smallest go to infinity. Inequality (6.37) shows that the smallest eigen-
value in the OLS case cannot exceed the finite upper bound γ OLS . This study
confirms the analysis in Section 6.1.1.1. Also, if b 2 = 0, which means that
b range ( A ) , the upper bound (6.37) implies that α n + 1 = 0 (a compatible system
of equations).
By looking again at the matrix K [see eq. (6.17)] and identifying the principal
(scalar) submatrix b T b / 2 ( 1 ζ ) , by means of a consequence of the interlacing
theorem [199, Prob. 2, p. 225], we obtain
ζ
b T b
α n + 1
ζ ) α 1
(6.39)
2
(
1
Hence, defining
0
OLS
1
2 b T b
γ
= E OLS (
) =
0
it follows that
0
OLS
γ OLS
1
γ
α n + 1
α 1
(6.40)
ζ
1
ζ
0
OLS
It can be concluded that
γ
/(
1
ζ)
is a better lower bound for
α
1 , while
γ
/(
1
ζ)
is a better upper bound for
α
1 .
OLS
n
+
6.2.1 Note on the Taylor Approximation of the Bounds of the
Eigenvalues of K
The bounds for the eigenvalues of matrix K expressed by (6.38) can be approx-
imated for ζ [0, 1] by a Taylor expansion of the first order:
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